1. ## Rings and Ideals

Need some help...

Let $\displaystyle \mathbb{Z}[\sqrt{-5}]$ the ring of gaussian integers.

1)Show that $\displaystyle <3>=<3;1+\sqrt{-5}><3;1-\sqrt{-5}>$
($\displaystyle <a>$ means the ideal generated by $\displaystyle a$)

2)Prove that the ideal on the right is prime.

Thank you folks!

2. Originally Posted by Inti

Let $\displaystyle \mathbb{Z}[\sqrt{-5}]$ the ring of gaussian integers.

1) Show that $\displaystyle <3>=<3,1+\sqrt{-5}><3,1-\sqrt{-5}>$
Hint: we have $\displaystyle 3=3^2 - (1+\sqrt{-5})(1-\sqrt{-5}).$ thus: $\displaystyle <3> \subseteq <3,1+\sqrt{-5}><3,1-\sqrt{-5}>.$ for the other side of the inclusion, note that $\displaystyle (1+\sqrt{-5})(1-\sqrt{-5})=6 \in <3>.$

2) Prove that the ideals on the right is prime.
Hint: define the map $\displaystyle f: \mathbb{Z}[\sqrt{-5}] \longrightarrow \mathbb{Z}_3$ by $\displaystyle f(a+b\sqrt{-5})=a-b.$ show that $\displaystyle f$ is a well-defined surjective ring homomorphism and: $\displaystyle \ker f = <3, 1+\sqrt{-5}>.$

3. there´s one thing that´s not clear to me:
I´ve just proved that $\displaystyle \ker f = <3,1+\sqrt{-5}>$, which means that $\displaystyle \mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}>$ is isomorphic to $\displaystyle \mathbb{Z}3$, which is a field.
That allows me to say that the ideal is prime?
As far as I´m concerned, the quotient of a ring by a prime ideal is a domain.
I´m kind of confused with that...
But the help was very helpful

And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??

Thanks!

4. Originally Posted by Inti

there´s one thing that´s not clear to me:
I´ve just proved that $\displaystyle \ker f = <3,1+\sqrt{-5}>$, which means that $\displaystyle \mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}>$ is isomorphic to $\displaystyle \mathbb{Z}3$, which is a field.
That allows me to say that the ideal is prime?
that means the ideal is maximal and we know that every maximal ideal is prime.

And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??
no. if $\displaystyle I$ and $\displaystyle J$ are two ideals, then $\displaystyle IJ=\{\sum_{k=1}^na_kb_k: \ n \in \mathbb{N}, \ a_k \in I, b_k \in J \}.$ however, for example, if you want to prove that $\displaystyle IJ \subseteq K,$ where $\displaystyle K$ is an ideal, then you only need to prove that

$\displaystyle ab \in K$ for all $\displaystyle a \in I, b \in J.$